Nilpotent Element is Contained in Prime Ideals
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Theorem
Let $R$ be a commutative ring with unity.
Let $\mathfrak p \subset R$ be a prime ideal of $R$.
Let $x \in R$ be a nilpotent element of $R$.
Then $x \in \mathfrak p$.
Proof
Let $x$ be nilpotent in $R$ as asserted.
Then by definition:
- $\exists n \in \Z_{>0}: x^n = 0$
But $0 \in \mathfrak p$ so:
- $x^n \in \mathfrak p$
\(\ds x\) | \(\in\) | \(\ds \map \Rad {\mathfrak p}\) | Definition 1 of Radical of Ideal of Ring | |||||||||||
\(\ds \) | \(=\) | \(\ds \mathfrak p\) | Radical of Prime Ideal |
$\blacksquare$